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CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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Ell-one and ell-infinity are not reflexive

Statement refuted

Assume AC. The classical sequence spaces 1 and are reflexive.

Facts & Assumptions

[A1]
[L1]

Under the ultrafilter lemma, DC, and Hahn--Banach, real and complex 1 are not reflexive (Ell one is not reflexive).

[L2]

Under Hahn--Banach and Countable Choice, X is reflexive exactly when X is reflexive (A Banach space is reflexive if and only if its dual is reflexive).

[L3]

The real dual of 1 is (Counting measure specializes the representation theorem to p and q), and the same holds over the complex field (The complex continuous dual of ell-one is ell-infinity).

[L4]

The dual () is isometrically ba, and under AC the countably additive charges form its proper 1 subspace (The dual of ell-infinity is ba, The countably additive part of ba is ell-one).

Counterexample

technique · counterexample

Given: The objects and hypotheses in the Statement.

1.1

AC in [A1] supplies the ultrafilter lemma, DC, Hahn--Banach, and Countable [given, A1, L1, L2] Choice needed by [L1] and [L2]. Thus [L1] already refutes reflexivity of 1, over both scalar fields.

A1L1
2.1

By [L3], (1)=. If were reflexive, [given, L3, L2, step 1.1] the reverse implication in [L2] would make 1 reflexive, contradicting step 1.1. Hence is not reflexive.

L2L3step 1.1
3.1

Independently, [L4] exhibits the bidual surplus: under the identification [given, L4, A1, step 2.1] (1)=()=ba, the canonical 1 image is only the proper subspace of countably additive charges. This is a concrete failed- surjectivity witness consistent with steps 1.1-2.1.

A1L4

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources